Define $p(a,b,c)=c(a+b)^2-a(b+c)-b(c+a)$, so that you seek positive $a,b,c$ for which $p(a,b,c)=0$. There is the solution $(1,1,1)$, but not others. First note that
$$p(a,b,1)=a(a-1)+b(b-1)$$ which is positive unless $a=b=1$ (going with our solution $(1,1,1)$.
Now write $p(a,b,c)$ in the form
Here the first term is positive, and the third is nonnegative, so that provided we show that $c \ge 2$ leads to the term in square brackets being nonnegative, we can conclude $p(a,b,c)>0$ whenever $c \ge 2$.
So to finish, we have when $c$ is at least 2 that
$$2bc-2b-c=c(2b-1)-2b \\ \ge 2(2b-1)-2b=2(b-1)\ge 0.$$